\section{Rodrigues' Formula}

Here we will show how to derive \textit{Rodrigues' Formula}. First we wanna prove two prerequisite formula regarding skew symmetric matrix,

\begin{equation}\nonumber
\begin{split}
\hat{\textbf{w}}^2 &= \textbf{ww}^T - \|\textbf{w}\|^2\textbf{I}\\
\hat{\textbf{w}}^3 &= - \|\textbf{w}\|^2\hat{\textbf{w}}
\end{split}
\end{equation}

We will show the first one using tensor algebra and then the second one can be got directly.

\begin{equation}\nonumber
\begin{split}
[\hat{\textbf{w}}^2]_{ij} &= \hat{\textbf{w}}_{ik}\hat{\textbf{w}}_{kj} = \boldsymbol{\epsilon}_{ikm}\textbf{w}_m\boldsymbol{\epsilon}_{kjn}\textbf{w}_n\\
&=- \boldsymbol{\epsilon}_{kim}\boldsymbol{\epsilon}_{kjn}\textbf{w}_m\textbf{w}_n = (\boldsymbol{\delta}_{in}\boldsymbol{\delta}_{mj}-\boldsymbol{\delta}_{ij}\boldsymbol{\delta}_{mn})\textbf{w}_m\textbf{w}_n\\
&=\boldsymbol{\delta}_{in}\boldsymbol{\delta}_{mj}\textbf{w}_m\textbf{w}_n - \boldsymbol{\delta}_{ij}\boldsymbol{\delta}_{mn}\textbf{w}_m\textbf{w}_n\\
&=\textbf{w}_i\textbf{w}_j - \boldsymbol{\delta}_{ij}\textbf{w}_l\textbf{w}_l\\
&=[\textbf{ww}^T]_{ij} - [\|\textbf{w}\|^2\textbf{I}]_{ij} = [\textbf{ww}^T - \|\textbf{w}\|^2\textbf{I}]_{ij}
\end{split}
\end{equation}

Then the second one is easy to show by simply multiplying both sides of the first equation with $\hat{\textbf{w}}$. With the above two formula, we start to derive $\textit{Rodrigues' Formula}$ which is the closed form for the exponential map of skew symmetric matrix. Now suppose we are give a unit vector $\textbf{w}$ and a real number $\theta$, we want to know what $e^{\hat{\textbf{w}}\theta}$ is. Here is how it works

\begin{equation}\nonumber
\begin{split}
e^{\hat{\textbf{w}}\theta} &= \textbf{I} + \theta\hat{\textbf{w}} + \frac{\theta^2}{2!}\hat{\textbf{w}}^2 +  \frac{\theta^3}{3!}\hat{\textbf{w}}^3 + \cdots\\
&= \textbf{I} + (\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots)\hat{\textbf{w}} + (\frac{\theta^2}{2!} - \frac{\theta^4}{4!} + \frac{\theta^6}{6!} - \cdots)\hat{\textbf{w}}^2\\
&= \textbf{I} + \hat{\textbf{w}}\sin\theta + \hat{\textbf{w}}^2(1 - \cos\theta)
\end{split}
\end{equation}

Actually it's easy to show that $e^{\hat{\textbf{a}}}$ is O(3). Moreover, due to the continuity of the determinant mapping and exponential mapping as well as the fact that $det(e^{\hat{\textbf{0}}}) = 1$, $e^{\hat{\textbf{a}}}$ is actually SO(3). An important conclusion is that exponential operation is actually a surjective mapping from so(3) to SO(3).